CAT Probability Combinatorics Questions & Solutions

A sample of real CAT Probability Combinatorics past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 49 Probability Combinatorics questions in all — sign up free to practise them timed.

Q1.CAT 2025
A cafeteria offers 5 types of sandwiches. Moreover, for each type of sandwich, a customer can choose one of 4 breads and opt for either small or large sized sandwich. Optionally, the customer may also add up to 2 out of 6 available sauces. The number of different ways in which an order can be placed for a sandwich, is
- 880
- 840
- 800
- 600
Show solution
Number of ways to choose a sandwich = $^5C_1$ ways

Number of ways to choose a bread = $^4C_1$ ways

Number of ways to choose bread size = $^2C_1$ ways

Number of ways to choose sauces = $^6C_0+^6C_1+^6C_2=1+6+15=22$ ways

So, number of different ways = $^5C_1\times\ ^4C_1\times\ ^2C_1\times\ 22=5\times\ 4\times\ 2\times\ 22=880$ ways.
Q2.CAT 2022
The arithmetic mean of all the distinct numbers that can be obtained by rearranging the digits in 1421, including itself, is
- 2222
- 2442
- 2592
- 3333
Show solution
The number of 4-digit numbers possible using 1,1,2, and 4 is $\frac{4!}{2!}=12$

Number of 1's, 2's and 4's in units digits will be in the ratio 2:1:1, i.e. 6 1's, 3 2's and 3 4's.

Sum = 6(1) + 3(2) + 3(4) = 24

Similarly, in tens digit, hundreds digit and thousands digit as well.

Therefore, sum = 24 + 24(10) + 24(100) + 24(1000) = 24(1111)

Mean = $\frac{24\left(1111\right)}{12}=2222$

The answer is option A.
Q3.CAT 2021
A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is
Answer: 50
Show solution
The question asks for the number of 4 digit numbers using only the digits 1, 2, and 3 such that the digits 2 and 3 appear at least once.

The different possibilities include :

Case 1:The four digits are ( 2, 2, 2, 3). Since the number 2 is repeated 3 times. The total number of arrangements are :

$\frac{4!}{3!}$ = 4.

Case 2: The four digits are 2, 2, 3, 3. The total number of four-digit numbers formed using this are :

$\frac{4!}{2!\cdot2!}=\ 6$

Case 3: The four digits are 2, 3, 3, 3. The number of possible 4 digit numbers are :

$\frac{4!}{3!}$ = 4

Case4: The four digits are 2, 3, 3, 1. The number of possible 4 digit numbers are :

$\frac{4!}{2!}=\ 12$

Case5: Using the digits 2, 2, 3, 1. The number of possible 4 digit numbers are :

$\frac{4!}{2!}=\ 12$

Case 6: Using the digits 2, 3, 1, 1. The number of possible 4 digit numbers are :

$\frac{4!}{2!}=\ 12$

A total of 12 + 12 + 12 + 4 + 6 + 4 = 50 possibilities.

Alternatively

We have to form 4 digit numbers using 1,2,3 such that 2,3 appears at least once
So the possible cases :

Now we get $\frac{4!}{2!}\times\ 3$ = 36 ( When one digit is used twice and the remaining two once )
$\frac{4!}{3!}\times\ 2$ = 8 ( When 1 is used 0 times and 2 and 3 is used 3 times or 1 time )
$\frac{4!}{2!\times\ 2!}=\ 6$ ( When 2 and 3 is used 2 times each )
So total numbers = 36+8+6 =50
Q4.CAT 2018
In a tournament, there are 43 junior level and 51 senior level participants. Each pair of juniors play one match. Each pair of seniors play one match. There is no junior versus senior match. The number of girl versus girl matches in junior level is 153, while the number of boy versus boy matches in senior level is 276. The number of matches a boy plays against a girl is
Answer: 1098
Show solution
In a tournament, there are 43 junior level and 51 senior level participants.

Let 'n' be the number of girls on junior level. It is given that the number of girl versus girl matches in junior level is 153.

$\Rightarrow$ nC2 = 153

$\Rightarrow$ n(n-1)/2 = 153

$\Rightarrow$ n(n-1) = 306

=> n $^{2}$ -n-306 = 0

=> (n+17)(n-18)=0

=> n=18 (rejecting n=-17)

Therefore, number of boys on junior level = 43 - 18 = 25.

Let 'm' be the number of boys on senior level. It is given that the number of boy versus boy matches in senior level is 276.

$\Rightarrow$ mC2 = 276

$\Rightarrow$ m = 24

Therefore, number of girls on senior level = 51 - 24 = 27.

Hence, the number of matches a boy plays against a girl = 18*25+24*27 = 1098
Q5.CAT 2017
In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?
Answer: 6
Show solution
After Amal, Bimal and Kamal are given their minimum required pens, the pens left are 8 - (1 + 2 + 3) = 2 pens
Now these two pens have to be divided between three persons so that each person can get zero pens = $^{2+3-1}C_{3-1}$ = $^4C_2$ = 6
Passage
Directions for the next two questions: The figure below shows the plan of a town. The streets are at right angles to each other. A rectangular park (P) is situated inside the town with a diagonal road running through it. There is also a prohibited region (D) in the town.

Q6.CAT 2008
Neelam rides her bicycle from her house at A to her office at B, taking the shortest path. Then the number of possible shortest paths that she can choose is

[CAT 2008]
- 60
- 75
- 45
- 90
- 72
Show solution

The shortest route from A to B is via the diagonal OQ in the square P. One can travel from A to O in 4!/2!*2! ways. The shortest way from O to Q is through the diagonal only.From Q to B can be travelled in 6!/4!*2! ways.

The total number of ways is, therefore, (4!/2!*2!) * (6!/4!*2!) = 6*15 = 90
Q7.CAT 2005
In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls, and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is
- 200
- 216
- 235
- 256
Show solution
Number of games in which both the players are girls = $^GC_2$ where G is the number of girls
$^GC_2 = 45$
$^{10}C_2 = 45$
So, G = 10
Similarly, number of games in which both the players are boys = $^BC_2$ , where B is the number of boys
$^BC_2 = 190$
$^{20}C_2 = 190$
So, B = 20
So, number of games in which one player is a boy and the other player is a girl is 20*10 = 200
Q8.CAT 2004
In the adjoining figure, the lines represent one-way roads allowing travel only northwards or only westwards. Along how many distinct routes can a car reach point B from point A?
- 15
- 35
- 120
- 336
Show solution
The person has to take 3 steps north and 4 steps west, in whatever way he travels.

Total steps = 7, 3 north and 4 west.

Number of ways = 7!/(4!3!) = 35
Q9.CAT 2003
A graph may be defined as a set of points connected by lines called edges. Every edge connects a pair of points. Thus, a triangle is a graph with 3 edges and 3 points. The degree of a point is the number of edges connected to it. For example, a triangle is a graph with three points of degree 2 each. Consider a graph with 12 points. It is possible to reach any point from any point through a sequence of edges. The number of edges, e, in the graph must satisfy the condition
- $11 \leq e \leq 66$
- $10 \leq e \leq 66$
- $11 \leq e \leq 65$
- $0 \leq e \leq 11$
Show solution
Take any 12 points.
The maximum number of edges which can be drawn through these 12 points are $^{12}C_2$ = 66
The minimum number of edges which can be drawn through these 12 points are 12-1 = 11 as the resulting figure need not be closed. It might be open.
Q10.CAT 2002
In how many ways is it possible to choose a white square and a black square on a chessboard so that the squares must not lie in the same row or column?
- 56
- 896
- 60
- 768
Show solution
First a black square can be selected in 32 ways. Out of remaining rows and columns, 24 white squares remain. 1 white square can them be chosen in 24 ways. So total no. of ways of selection is 32*24 = 768.
Q11.CAT 2000
One red flag, three white flags and two blue flags are arranged in a line such that,

A. no two adjacent flags are of the same colour

B. the flags at the two ends of the line are of different colours.

In how many different ways can the flags be arranged?
- 6
- 4
- 10
- 2
Show solution
The three white flags can be arranged in the following two ways:

__ W __ W __ W or W __ W __ W __

In the blanks, the 2 blue and one red flag can be arranged in 3 ways.
So, the total number of arrangements is 2*3 = 6
Q12.CAT 1997
In how many ways can eight directors, the vice chairman and chairman of a firm be seated at a round table, if the chairman has to sit between the the vice chairman and a specific director?
- 9! × 2
- 2 × 8!
- 2 × 7!
- None of these
Show solution
Chariman, Vice-Chairman and the director can be made as a group such that Chairman sits between the Vice-Chairman and the director. This group can be formed in 2 ways.

Each of the remaining 7 directors and the group can be arranged in 7! ways.

=> Total number of ways = 2 * 7!.

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CAT Probability Combinatorics Questions & Solutions

Q1.CAT 2025

A cafeteria offers 5 types of sandwiches. Moreover, for each type of sandwich, a customer can choose one of 4 breads and opt for either small or large sized sandwich. Optionally, the customer may also add up to 2 out of 6 available sauces. The number of different ways in which an order can be placed for a sandwich, is

Show solution

Number of ways to choose a sandwich = $^5C_1$ ways

Number of ways to choose a bread = $^4C_1$ ways

Number of ways to choose bread size = $^2C_1$ ways

Number of ways to choose sauces = $^6C_0+^6C_1+^6C_2=1+6+15=22$ ways

So, number of different ways = $^5C_1\times\ ^4C_1\times\ ^2C_1\times\ 22=5\times\ 4\times\ 2\times\ 22=880$ ways.

Q2.CAT 2022

The arithmetic mean of all the distinct numbers that can be obtained by rearranging the digits in 1421, including itself, is

2222
2442
2592
3333

Show solution

The number of 4-digit numbers possible using 1,1,2, and 4 is $\frac{4!}{2!}=12$

Number of 1's, 2's and 4's in units digits will be in the ratio 2:1:1, i.e. 6 1's, 3 2's and 3 4's.

Sum = 6(1) + 3(2) + 3(4) = 24

Similarly, in tens digit, hundreds digit and thousands digit as well.

Therefore, sum = 24 + 24(10) + 24(100) + 24(1000) = 24(1111)

Mean = $\frac{24\left(1111\right)}{12}=2222$

The answer is option A.

Q3.CAT 2021

A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is

Answer: 50

Show solution

The question asks for the number of 4 digit numbers using only the digits 1, 2, and 3 such that the digits 2 and 3 appear at least once.

The different possibilities include :

Case 1:The four digits are ( 2, 2, 2, 3). Since the number 2 is repeated 3 times. The total number of arrangements are :

$\frac{4!}{3!}$ = 4.

Case 2: The four digits are 2, 2, 3, 3. The total number of four-digit numbers formed using this are :

$\frac{4!}{2!\cdot2!}=\ 6$

Case 3: The four digits are 2, 3, 3, 3. The number of possible 4 digit numbers are :

$\frac{4!}{3!}$ = 4

Case4: The four digits are 2, 3, 3, 1. The number of possible 4 digit numbers are :

$\frac{4!}{2!}=\ 12$

Case5: Using the digits 2, 2, 3, 1. The number of possible 4 digit numbers are :

$\frac{4!}{2!}=\ 12$

Case 6: Using the digits 2, 3, 1, 1. The number of possible 4 digit numbers are :

$\frac{4!}{2!}=\ 12$

A total of 12 + 12 + 12 + 4 + 6 + 4 = 50 possibilities.

Alternatively

We have to form 4 digit numbers using 1,2,3 such that 2,3 appears at least once
So the possible cases :

Now we get $\frac{4!}{2!}\times\ 3$ = 36 ( When one digit is used twice and the remaining two once )
$\frac{4!}{3!}\times\ 2$ = 8 ( When 1 is used 0 times and 2 and 3 is used 3 times or 1 time )
$\frac{4!}{2!\times\ 2!}=\ 6$ ( When 2 and 3 is used 2 times each )
So total numbers = 36+8+6 =50

Q4.CAT 2018

In a tournament, there are 43 junior level and 51 senior level participants. Each pair of juniors play one match. Each pair of seniors play one match. There is no junior versus senior match. The number of girl versus girl matches in junior level is 153, while the number of boy versus boy matches in senior level is 276. The number of matches a boy plays against a girl is

Answer: 1098

Show solution

In a tournament, there are 43 junior level and 51 senior level participants.

Let 'n' be the number of girls on junior level. It is given that the number of girl versus girl matches in junior level is 153.

$\Rightarrow$ nC2 = 153

$\Rightarrow$ n(n-1)/2 = 153

$\Rightarrow$ n(n-1) = 306

=> n $^{2}$ -n-306 = 0

=> (n+17)(n-18)=0

=> n=18 (rejecting n=-17)

Therefore, number of boys on junior level = 43 - 18 = 25.

Let 'm' be the number of boys on senior level. It is given that the number of boy versus boy matches in senior level is 276.

$\Rightarrow$ mC2 = 276

$\Rightarrow$ m = 24

Therefore, number of girls on senior level = 51 - 24 = 27.

Hence, the number of matches a boy plays against a girl = 18*25+24*27 = 1098

Q5.CAT 2017

In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?

Answer: 6

Show solution

After Amal, Bimal and Kamal are given their minimum required pens, the pens left are 8 - (1 + 2 + 3) = 2 pens
Now these two pens have to be divided between three persons so that each person can get zero pens = $^{2+3-1}C_{3-1}$ = $^4C_2$ = 6

Passage

Directions for the next two questions: The figure below shows the plan of a town. The streets are at right angles to each other. A rectangular park (P) is situated inside the town with a diagonal road running through it. There is also a prohibited region (D) in the town.

Q6.CAT 2008

Neelam rides her bicycle from her house at A to her office at B, taking the shortest path. Then the number of possible shortest paths that she can choose is

[CAT 2008]

Show solution

The shortest route from A to B is via the diagonal OQ in the square P. One can travel from A to O in 4!/2!*2! ways. The shortest way from O to Q is through the diagonal only.From Q to B can be travelled in 6!/4!*2! ways.

The total number of ways is, therefore, (4!/2!*2!) * (6!/4!*2!) = 6*15 = 90

Q7.CAT 2005

In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls, and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is

Show solution

Number of games in which both the players are girls = $^GC_2$ where G is the number of girls
$^GC_2 = 45$
$^{10}C_2 = 45$
So, G = 10
Similarly, number of games in which both the players are boys = $^BC_2$ , where B is the number of boys
$^BC_2 = 190$
$^{20}C_2 = 190$
So, B = 20
So, number of games in which one player is a boy and the other player is a girl is 20*10 = 200

Q8.CAT 2004

In the adjoining figure, the lines represent one-way roads allowing travel only northwards or only westwards. Along how many distinct routes can a car reach point B from point A?

Show solution

The person has to take 3 steps north and 4 steps west, in whatever way he travels.

Total steps = 7, 3 north and 4 west.

Number of ways = 7!/(4!3!) = 35

Q9.CAT 2003

A graph may be defined as a set of points connected by lines called edges. Every edge connects a pair of points. Thus, a triangle is a graph with 3 edges and 3 points. The degree of a point is the number of edges connected to it. For example, a triangle is a graph with three points of degree 2 each. Consider a graph with 12 points. It is possible to reach any point from any point through a sequence of edges. The number of edges, e, in the graph must satisfy the condition

$11 \leq e \leq 66$
$10 \leq e \leq 66$
$11 \leq e \leq 65$
$0 \leq e \leq 11$

Show solution

Take any 12 points.

The maximum number of edges which can be drawn through these 12 points are $^{12}C_2$ = 66

The minimum number of edges which can be drawn through these 12 points are 12-1 = 11 as the resulting figure need not be closed. It might be open.

Q10.CAT 2002

In how many ways is it possible to choose a white square and a black square on a chessboard so that the squares must not lie in the same row or column?

Show solution

First a black square can be selected in 32 ways. Out of remaining rows and columns, 24 white squares remain. 1 white square can them be chosen in 24 ways. So total no. of ways of selection is 32*24 = 768.

Q11.CAT 2000

One red flag, three white flags and two blue flags are arranged in a line such that,

A. no two adjacent flags are of the same colour

B. the flags at the two ends of the line are of different colours.

In how many different ways can the flags be arranged?

Show solution

The three white flags can be arranged in the following two ways:

__ W __ W __ W or W __ W __ W __

In the blanks, the 2 blue and one red flag can be arranged in 3 ways.
So, the total number of arrangements is 2*3 = 6

Q12.CAT 1997

In how many ways can eight directors, the vice chairman and chairman of a firm be seated at a round table, if the chairman has to sit between the the vice chairman and a specific director?

9! × 2
2 × 8!
2 × 7!
None of these

Show solution

Chariman, Vice-Chairman and the director can be made as a group such that Chairman sits between the Vice-Chairman and the director. This group can be formed in 2 ways.

Each of the remaining 7 directors and the group can be arranged in 7! ways.

=> Total number of ways = 2 * 7!.

37+ more Probability Combinatorics questions inside

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CAT Probability Combinatorics previous year questions with solutions — AthenaPrep